Let T: P2(R) R3 Be Defined As T(p(x))=(p(-1),p(0),p(1)) A)Show That T Is Linear B)Find Ker(T) C)Is T

Let T: P2(R) R3 Be Defined As T(p(x))=(p(-1),p(0),p(1)) A)Show That T Is Linear B)Find Ker(T) C)Is T

Introduction

In the study of linear algebra, understanding the properties of linear transformations is fundamental. The transformation T: P2(R) → R^3, defined by T(p(x)) = (p(-1), p(0), p(1)), where P2(R) is the space of all real polynomials of degree at most 2, offers a rich example to explore these properties. This article aims to demonstrate the linearity of T, determine its kernel (Ker(T)), and analyze whether T is an injective, surjective, or bijective transformation. These concepts are essential for understanding the structure of polynomial spaces and their mappings into coordinate spaces.

Part A: Showing That T Is Linear

Definition of Linearity

A transformation T: V → W between two vector spaces V and W over a field (in this case, ℝ) is linear if it satisfies two properties for all vectors u, v ∈ V and all scalars c ∈ ℝ:

• Additivity: T(u + v) = T(u) + T(v)
• Homogeneity: T(cu) = cT(u)

Applying the Definition to T

Let p(x) and q(x) be any polynomials in P2(R), and c be any real scalar.

• To verify additivity:

T(p(x) + q(x)) = ((p + q)(-1), (p + q)(0), (p + q)(1))

= (p(-1) + q(-1), p(0) + q(0), p(1) + q(1))

= (p(-1), p(0), p(1)) + (q(-1), q(0), q(1))

= T(p(x)) + T(q(x))

• To verify homogeneity:

T(c p(x)) = (c p(-1), c p(0), c p(1))

= c (p(-1), p(0), p(1))

= c T(p(x))

Since both properties hold for any p(x), q(x) ∈ P2(R) and c ∈ ℝ, T is a linear transformation.

Part B: Finding the Kernel of T (Ker(T))

Definition of Kernel

The kernel of T, denoted as Ker(T), consists of all polynomials p(x) ∈ P2(R) such that T(p(x)) = (0, 0, 0). In other words, Ker(T) is the set of all polynomials that map to the zero vector in ℝ^3.

Finding Polynomials in Ker(T)

Suppose p(x) = a x^2 + b x + c ∈ P2(R). Then,

T(p(x)) = (p(-1), p(0), p(1)) = (0, 0, 0)

This leads to the system of equations:

1. p(-1) = a(-1)^2 + b(-1) + c = a - b + c = 0


2. p(0) = a(0)^2 + b(0) + c = c = 0


3. p(1) = a(1)^2 + b(1) + c = a + b + c = 0

From equation (2), c = 0.

Substitute c = 0 into equations (1) and (3):

• a - b = 0 → a = b


• a + b = 0 → a + a = 0 → 2a = 0 → a = 0

Since a = 0, then b = 0 (from a = b). Therefore, the polynomial p(x) reduces to:

p(x) = a x^2 + b x + c = 0 x^2 + 0 x + 0 = 0

The only polynomial in Ker(T) is the zero polynomial.

Conclusion on Ker(T)

Ker(T) = { p(x) ∈ P2(R) | p(x) ≡ 0 } = { 0 }

This indicates that the transformation T is injective because its kernel contains only the zero polynomial.

Part C: Is T

Analysis of T’s Properties

Given the previous results, T is a linear transformation from P2(R) to ℝ^3 with the following properties:

• Injectivity: Since Ker(T) = {0}, T is injective. No non-zero polynomial maps to the zero vector, meaning T is one-to-one.
• Surjectivity: To determine if T is onto, we need to check whether every vector in ℝ^3 has a pre-image in P2(R).

Is T Surjective?

For T to be surjective, for any (x, y, z) ∈ ℝ^3, there must exist a polynomial p(x) = a x^2 + b x + c such that:

p(-1) = x

p(0) = y

p(1) = z

This translates into the system:

1. a - b + c = x


2. c = y


3. a + b + c = z

Substituting c = y into the other equations:

• a - b + y = x → a - b = x - y


• a + b + y = z → a + b = z - y

Adding these two equations:

(a - b) + (a + b) = (x - y) + (z - y)

2a = x - y + z - y = x + z - 2y

Thus,

a = (x + z - 2y)/2

Using a in the equation a + b = z - y:

((x + z - 2y)/2) + b = z - y

b = z - y - (x + z - 2y)/2

Simplify b:

b = (2(z - y) - (x + z - 2y)) / 2

b = (2z - 2y - x - z + 2y) / 2

b = (z - x) / 2

Similarly, c = y.

Since for arbitrary (x, y, z), we can find real numbers a, b, c satisfying these equations, T is surjective.

Conclusion on T

Because T is both injective and surjective from P2(R) to ℝ^3, T is a bijective linear transformation (an isomorphism).

Summary of Results

• Linearity: T is linear because it preserves addition and scalar multiplication.
• Kernel: Ker(T) = {0}, meaning T is injective.
• Surjectivity: T is onto ℝ^3, making it surjective.
• Overall: T is an isomorphism between P2(R) and ℝ^3.

Implications and Applications

Understanding the properties of T provides insights into how polynomial evaluations at specific points can serve as a basis for mapping polynomial spaces into coordinate spaces. Such transformations are fundamental in various applications, including polynomial interpolation, numerical analysis, and computer graphics. The fact that T is an isomorphism highlights that evaluating polynomials at three distinct points completely characterizes the polynomial, which is a cornerstone in polynomial interpolation techniques like Lagrange interpolation.

Conclusion

In this comprehensive analysis, we demonstrated that the linear transformation T, defined as evaluation at points -1, 0, and 1, is a bijective linear mapping from the space of quadratic polynomials to ℝ^3. Its kernel contains only the zero polynomial, confirming injectivity, and the ability to solve for any vector in ℝ^3 indicates surjectivity. These properties make T a powerful example of an isomorphism in linear algebra, illustrating the deep connection between polynomial spaces and coordinate representations. Understanding such transformations enhances our grasp of fundamental linear algebra concepts and their practical applications across mathematics and engineering disciplines.